MHTCET 22 Apr 2025 Shift 1 Paper

Concept:The percentage error in $g$ is found using error propagation in the pendulum formula $T = 2\pi\sqrt{\frac{l}{g}}$.Explanation:From $T = 2\pi\sqrt{\frac{l}{g}}$, we derive $g = \frac{4\pi^2 l}{T^2}$.The relative error in $g$ combines the relative error in $l$ and twice the relative error in $T$:$$\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}$$Here, $l = 100\ \text{cm} = 1.00\ \text{m}$ and $\Delta l = 1\ \text{mm} = 0.001\ \text{m}$.So, $\frac{\Delta l}{l} = \frac{0.001}{1.00} = 0.001 = 0.1\%$.For 100 oscillations, total measured time is $100 \times 2 = 200\ \text{s}$.The stopwatch least count is $0.1\ \text{s}$, giving total time error $\Delta t = 0.1\ \text{s}$.Thus, $\frac{\Delta T}{T} = \frac{\Delta t}{t} = \frac{0.1}{200} = 0.0005 = 0.05\%$.The contribution from the time term is $2 \times 0.05\% = 0.1\%$.Total percentage error in $g$ is $0.1\% + 0.1\% = 0.2\%$.Answer:The percentage error in gravitational acceleration $g$ is $0.2\%$, which matches option A.